Question

Consider a collision between an x-ray photon of initial energy $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{e}\mathit{V}$and an electron at rest, in which the photon is scattered backward and the electron is knocked forward.

(a) What is the energy of the backscattered photon?

(b) What is the kinetic energy of the electron?

Short answer

(a) The energy of the backscattered photon is$41.8\text{\hspace{0.17em}keV}$.

(b) The kinetic energy of the electron is $8.2\text{\hspace{0.17em}keV}$.

Step-by-step solution

Write the given data from the question.

Initial energy of the photon,

Determine the formulas to calculate the energy of the backscattered photon and kinetic energy of the electron. 

The expression to calculate the initial energy is given as follows.

…(i)

Here,h is the plank’s constant and c is the speed of light.

The expression to calculate the change in the wavelength is given as follows.

$\mathit{\Delta }\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{c}}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{)}$ …(ii)

Here,m is the mass of the electron.

The expression to calculate the kinetic energy of the electron is given as follows.

$\mathit{K}\mathbf{=}\mathit{E}\mathbf{-}{\mathit{E}}^{\mathbf{\text{'}}}$ …(iii)

Here, $E^{\text{'}}$is the energy of backscattered photon.

Calculate the energy of the backscattered photon.

The value of plank’s constant is, $6.62\times {10}^{-34}\text{\hspace{0.17em}mkg}/\text{s}$.

The value of mass of electron is,$9.11\times {10}^{-31}\text{\hspace{0.17em}kg}$.

Since the photon is backscattered and electron is knocked forward, therefore the angle, $\theta =180°$.

Substitute$180°$for$\theta$into equation (ii).

$\begin{array}{l}\Delta \lambda =\frac{h}{mc}(1-\cos 180°)\\ \Delta \lambda =\frac{h}{mc}(1-(-1))\\ \Delta \lambda =\frac{2h}{mc}\end{array}$

The new wavelength is given by,

role="math" localid="1663071812908" $\begin{array}{l}{\lambda }^{\text{'}}=\Delta \lambda +\lambda \\ {\lambda }^{\text{'}}=\lambda \left(\frac{\Delta \lambda }{\lambda }+1\right)\end{array}$

Calculate the energy of photon when it is backscattered.

$E^{\text{'}}=\frac{hc}{{\lambda }^{\text{'}}}$

Substitute $\lambda \left(\frac{\Delta \lambda }{\lambda }+1\right)$for${\lambda }^{\text{'}}$into above equation.

$\begin{array}{l}{E}^{\text{'}}=\frac{hc}{\lambda \left(\frac{\Delta \lambda }{\lambda }+1\right)}\\ E^{\text{'}}=\frac{\frac{hc}{\lambda }}{\left(\frac{\Delta \lambda }{\lambda }+1\right)}\end{array}$

Substitute E for$\frac{hc}{\lambda }$into above equation.

$E^{\text{'}}=\frac{E}{\left(\frac{\Delta \lambda }{\lambda }+1\right)}$

Substitute$2h/\text{mc}$for$\Delta \lambda$and $hc/E$for$\lambda$into above equation.

$\begin{array}{l}{E}^{\text{'}}=\frac{E}{\left(\frac{\frac{2h}{mc}}{\frac{hc}{E}}+1\right)}\\ E^{\text{'}}=\frac{E}{\left(\frac{2hE}{mh{c}^2}+1\right)}\\ E^{\text{'}}=\frac{E}{\left(\frac{2E}{m{c}^2}+1\right)}\end{array}$

And,

$\begin{array}{c}m{c}^2=9.11\times {10}^{-31}\text{\hspace{0.17em}kg}\times {(3\times {10}^8\text{\hspace{0.17em}m}/\text{s})}^2\\ =8.199\times {10}^{-14}\text{\hspace{0.17em}J}\\ =8.199\times {10}^{-14}\times 6.242\times {10}^{15}\text{\hspace{0.17em}keV}\\ =511.78\text{\hspace{0.17em}keV}\\ m{c}^2\simeq 511\text{\hspace{0.17em}keV}\end{array}$

Substitute$50\text{\hspace{0.17em}keV}$for E and $511\text{\hspace{0.17em}keV}$for $m{c}^2$into above equation

$\begin{array}{c}{E}^{\text{'}}=\frac{50\text{\hspace{0.17em}keV}}{\frac{2\times 50\text{\hspace{0.17em}keV}}{511\text{\hspace{0.17em}keV}}+1}\\ E^{\text{'}}=\frac{50\text{\hspace{0.17em}keV}}{\frac{100}{511}+1}\\ E^{\text{'}}=\frac{50\text{\hspace{0.17em}keV}}{1.1956}\\ E^{\text{'}}=41.8\text{\hspace{0.17em}keV}\end{array}$

Hence the energy of the backscattered photon is .$41.8\text{\hspace{0.17em}keV}$

Calculate the kinetic energy of the electron.

Calculate the kinetic energy of electron.

Substitute$41.8\text{\hspace{0.17em}keV}$for $E^{\text{'}}$and $50\text{\hspace{0.17em}keV}$for E into equation (iii).

$\begin{array}{l}K=50\text{\hspace{0.17em}keV}-41.8\text{\hspace{0.17em}keV}\\ K=8.2\text{\hspace{0.17em}keV}\end{array}$

Hence the kinetic energy of the electron is$8.2\text{\hspace{0.17em}keV}$.